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 {  "cells": [  {  "cell_type": "code",  "execution_count": null,  "metadata": {},  "outputs": [],  "source": []  },  {  "cell_type": "markdown",  "metadata": {},  "source": [  "# GCD / Euclidean algorithm\n",  "Explaination: https://brilliant.org/wiki/extended-euclidean-algorithm/"  ]  },  {  "cell_type": "code",  "execution_count": 1,  "metadata": {},  "outputs": [  {  "data": {  "text/plain": [  "5"  ]  },  "execution_count": 1,  "metadata": {},  "output_type": "execute_result"  }  ],  "source": [  "# Returns the greatest common divisor of a and b.\n",  "def gcd(a, b):\n",  " if b > a:\n",  " return gcd(b, a)\n",  " if a % b == 0:\n",  " return b\n",  " return gcd(b, a % b)\n",  "\n",  "gcd(35,30)"  ]  },  {  "cell_type": "markdown",  "metadata": {},  "source": [  "# Back substitution / Extended euclidean algorithm\n",  "Explaination: https://brilliant.org/wiki/extended-euclidean-algorithm/"  ]  },  {  "cell_type": "code",  "execution_count": 2,  "metadata": {},  "outputs": [  {  "name": "stdout",  "output_type": "stream",  "text": [  "ax + by = gcd(a,b) ==> 35*1 + 10*-3 = gcd(35,10)\n",  "\n",  "gcd(35,10) = 5\n",  "x = 1, y = -3\n"  ]  }  ],  "source": [  "def gcdExtended(a, b): \n",  " # Base Case \n",  " if a == 0 : \n",  " return b,0,1\n",  "\n",  " gcd,x1,y1 = gcdExtended(b%a, a)\n",  "\n",  " # Update x and y using results of recursive call \n",  " x = y1 - (b//a) * x1 \n",  " y = x1 \n",  "\n",  " return gcd, x, y \n",  "\n",  "a, b = 35,10\n",  "g, x, y = gcdExtended(a, b)\n",  "print(f\"ax + by = gcd(a,b) ==> {a}*{x} + {b}*{y} = gcd({a},{b})\\n\")\n",  "\n",  "print(f\"gcd({a},{b}) = {g}\")\n",  "print(f\"x = {x}, y = {y}\")"  ]  },  {  "cell_type": "markdown",  "metadata": {},  "source": [  "# Modular inverse, A^-1\n",  "Explanation: https://www.khanacademy.org/computing/computer-science/cryptography/modarithmetic/a/modular-inverses"  ]  },  {  "cell_type": "code",  "execution_count": 3,  "metadata": {},  "outputs": [  {  "name": "stdout",  "output_type": "stream",  "text": [  "3 * 5 ≡ 1 (mod 7)\n",  "Modular inverse for 3 (mod 7) is 5\n",  "a^-1 = 5\n"  ]  }  ],  "source": [  "# Naive method of finding modular inverse for A (mod C) \n",  "def inverse(a, c):\n",  " for b in range(c):\n",  " if (a*b)%c == 1: return b\n",  " print(\"NO INVERSE FOUND\")\n",  "\n",  "a,c = 3,7\n",  "b = inverse(a,c)\n",  "\n",  "print(f\"{a} * {b} ≡ 1 (mod {c})\")\n",  "print(f\"Modular inverse for {a} (mod {c}) is {b}\")\n",  "print(f\"a^-1 = {b}\")"  ]  },  {  "cell_type": "markdown",  "metadata": {},  "source": [  "# Chineese remainder theorem\n",  "Source: https://rosettacode.org/wiki/Chinese_remainder_theorem#Python"  ]  },  {  "cell_type": "code",  "execution_count": 4,  "metadata": {},  "outputs": [  {  "name": "stdout",  "output_type": "stream",  "text": [  "23\n"  ]  }  ],  "source": [  "from functools import reduce\n",  "def crt(n, a):\n",  " sum = 0\n",  " prod = reduce(lambda a, b: a*b, n)\n",  " for n_i, a_i in zip(n, a):\n",  " p = prod // n_i\n",  " sum += a_i * mul_inv(p, n_i) * p\n",  " return sum % prod\n",  "\n",  "def mul_inv(a, b):\n",  " b0 = b\n",  " x0, x1 = 0, 1\n",  " if b == 1: return 1\n",  " while a > 1:\n",  " q = a // b\n",  " a, b = b, a%b\n",  " x0, x1 = x1 - q * x0, x0\n",  " if x1 < 0: x1 += b0\n",  " return x1\n",  "\n",  "n = [3, 5, 7]\n",  "a = [2, 3, 2]\n",  "print(crt(n, a))"  ]  },  {  "cell_type": "markdown",  "metadata": {},  "source": [  "# Integer factorisation\n",  "Given an integer, find its prime factors"  ]  },  {  "cell_type": "code",  "execution_count": 5,  "metadata": {},  "outputs": [  {  "name": "stdout",  "output_type": "stream",  "text": [  "The factors of 641361 are [3, 7, 7, 4363], verified\n",  "Manual test: True\n"  ]  }  ],  "source": [  "# Brute force solution\n",  "def prime_factors(n):\n",  " i = 2\n",  " factors = []\n",  " while i * i <= n:\n",  " if n % i:\n",  " i += 1\n",  " else:\n",  " n //= i\n",  " factors.append(i)\n",  " if n > 1:\n",  " factors.append(n)\n",  " return factors\n",  "\n",  "# Check if the numbers in array a is the factors of n\n",  "def factor_test(a, n):\n",  " k = 1\n",  " for i in factors:\n",  " k=k*i\n",  " return k==n\n",  "\n",  "n = 641361\n",  "factors = prime_factors(n)\n",  "print(f\"The factors of {n} are {factors}, {'verified' if factor_test(factors, n) else 'ERROR'}\")\n",  "\n",  "print(\"Manual test:\", factor_test([3, 7, 7, 4363], 641361))"  ]  },  {  "cell_type": "markdown",  "metadata": {},  "source": [  "# Euler function φ and Z*n\n",  "From slides 2020-4135-l07 - Number Theory for Public Key Cryptography\n"  ]  },  {  "cell_type": "code",  "execution_count": 6,  "metadata": {},  "outputs": [  {  "name": "stdout",  "output_type": "stream",  "text": [  "φ(21) = 12 since the 12 items in Z*21 are each relatively prime to 21.\n",  "Z*21 = [1, 2, 4, 5, 8, 10, 11, 13, 16, 17, 19, 20]\n"  ]  }  ],  "source": [  "# Relatively prime means that gcd equals 1\n",  "def is_relative_prime(a,b):\n",  " if b == 0 or a == 0: return False\n",  " return gcd(a,b) == 1\n",  "\n",  "# The set of positive integers less than n and relatively prime \n",  "# to n form the reduced residue class Z∗n.\n",  "def z_star(n):\n",  " r = []\n",  " for i in range(1,n):\n",  " if is_relative_prime(i,n):\n",  " r.append(i)\n",  " return r\n",  "\n",  "n = 21\n",  "ar = z_star(n)\n",  "e = len(ar)\n",  "print(f\"φ({n}) = {e} since the {e} items in Z*{n} are each relatively prime to {n}.\")\n",  "print(f\"Z*{n} = {ar}\")"  ]  },  {  "cell_type": "markdown",  "metadata": {},  "source": [  "## Euler’s Totient\n",  "- https://www.dcode.fr/euler-totient\n",  "- https://www.ibnus.io/eulers-totient-function-using-python/\n",  "\n",  "In general:\n",  "\n",  "> $\\varphi(n) = n \\prod_{p \\mid n} \\left( 1 - \\frac{1}{p} \\right)$\n",  "\n",  "Examples:\n",  "> $\\varphi(16) = \\varphi(2^{4}) = 16*\\left(1-\\frac{1}{2}\\right) = 8$\n",  "\n",  "> $\\varphi(1125) = \\varphi(33555) = 1125\\left(1-\\frac{1}{3}\\right)\\left(1-\\frac{1}{5}\\right) = 600$\n"  ]  },  {  "cell_type": "code",  "execution_count": 7,  "metadata": {},  "outputs": [  {  "name": "stdout",  "output_type": "stream",  "text": [  "𝜑(1125) = 600\n"  ]  }  ],  "source": [  "# Euler's Totient Function\n",  "# using Euler's product formula\n",  "def totient(n) :\n",  " result = n # Initialize result as n\n",  "\n",  " # Consider all prime factors\n",  " # of n and for every prime\n",  " # factor p, multiply result with (1 – 1/p)\n",  " p = 2\n",  " while(p * p <= n) :\n",  "\n",  " # Check if p is a prime factor.\n",  " if (n % p == 0) :\n",  "\n",  " # If yes, then update n and result\n",  " while (n % p == 0) :\n",  " n = n // p\n",  " result = result * (1.0 - (1.0 / (float) (p)))\n",  " p = p + 1\n",  "\n",  " # If n has a prime factor\n",  " # greater than sqrt(n)\n",  " # (There can be at-most one\n",  " # such prime factor)\n",  " if (n > 1) :\n",  " result = result * (1.0 - (1.0 / (float)(n)))\n",  "\n",  " return (int)(result)\n",  "\n",  "n = 1125\n",  "print(f\"𝜑({n}) = {totient(n)}\")"  ]  },  {  "cell_type": "markdown",  "metadata": {},  "source": [  "## Finding a generator of Z*p\n",  "- https://crypto.stanford.edu/pbc/notes/numbertheory/gen.html\n",  "- Lecture 2, page 19\n",  "\n",  "A generator of $Z_p^∗$ is an element of order $p − 1$\n",  "\n",  "To find a generator of $Z_p^∗$ we can choose a value g and test it as follows:\n",  "1. Compute all the distinct prime factors of $p − 1$ and call them $f_1, f_2, ..., f_r$\n",  "2. Then $g$ is a generator as long as $g^{\\frac{p−1}{f_i}} \\neq 1 \\mod(p)$ for $i = 1,2,,...,r$"  ]  },  {  "cell_type": "code",  "execution_count": 8,  "metadata": {},  "outputs": [  {  "name": "stdout",  "output_type": "stream",  "text": [  "Is 3 a generator for Z*4? True\n"  ]  }  ],  "source": [  "# Finding a generator g of Z∗p\n",  "def is_generator(p, g):\n",  " pf = prime_factors(p-1)\n",  " for f in pf:\n",  " if (g**((p-1)/f))%p == 1:\n",  " return False\n",  " return True\n",  "\n",  "g = 3 # Generator\n",  "p = 4 # Z*p\n",  "\n",  "print(f\"Is {g} a generator for Z*{p}? {is_generator(p, g)}\")"  ]  },  {  "cell_type": "markdown",  "metadata": {},  "source": [  "# Testing for primality\n",  "From slides 2020-4135-l07 - Number Theory for Public Key Cryptography"  ]  },  {  "cell_type": "code",  "execution_count": 9,  "metadata": {},  "outputs": [  {  "name": "stdout",  "output_type": "stream",  "text": [  "137 True\n"  ]  }  ],  "source": [  "from math import sqrt\n",  "# Inefficient, but accurate method\n",  "def is_prime(n):\n",  " if n % 2 == 0 and n > 2: \n",  " return False\n",  " return all(n % i for i in range(3, int(sqrt(n)) + 1, 2))\n",  "\n",  "print(137, is_prime(137))"  ]  },  {  "cell_type": "code",  "execution_count": 10,  "metadata": {},  "outputs": [  {  "name": "stdout",  "output_type": "stream",  "text": [  "1105 is probable prime\n"  ]  }  ],  "source": [  "# Fermat primality test\n",  "\n",  "# Inputs: \n",  "# n: a value to test for primality;\n",  "# k: a parameter that determines the number of \n",  "# times to test for primality\n",  "from random import randint\n",  "def fermat_prime(n, k):\n",  " for _ in range(k):\n",  " a = randint(2,n-2)\n",  " # Fermat’s theorem says that if a number p is prime then \n",  " # a^p−1 mod p = 1 for all a with gcd(a,p) = 1.\n",  " if not (a**(n-1))%n == 1:\n",  " return \"composite\"\n",  " return \"probable prime\"\n",  "\n",  "n = 1105 # value to test for primality;\n",  "k = 1 # number of times to test for primality\n",  "\n",  "print(f\"{n} is {fermat_prime(n,k)}\")\n",  "\n",  "# First few Carmichael numbers are: 561, 1105, 1729, 2465 \n",  "# Will output probable prime for every a with gcd(a, n) = 1"  ]  },  {  "cell_type": "markdown",  "metadata": {},  "source": [  "## Miller-Rabin\n",  "From https://en.wikipedia.org/wiki/Miller%E2%80%93Rabin_primality_test \n",  "This test is not fooled by Charmichael numbers.\n",  "\n",  "To test n, you choose different a values, and test if \n",  "\n",  "> $a^{d*2^i} \\equiv 1 \\pmod{n}$ for $i = (0, 1, 2, ... r)$\n",  "\n",  "If this happens, look at the congruence on the calculation just before you got 1. If that value is not -1 (aka n-1), the value is composite aka not prime."  ]  },  {  "cell_type": "code",  "execution_count": 11,  "metadata": {},  "outputs": [  {  "name": "stdout",  "output_type": "stream",  "text": [  "168 = 21*2^3 = 168\n",  "65536 = 1*2^16 = 65536\n",  "\n",  "16127 is probably prime.\n",  "561 is composite.\n",  "2465 is composite.\n",  "65537 is probably prime.\n"  ]  }  ],  "source": [  "# Miller-Rabin primality test\n",  "# This test is not fooled by Charmichael numbers.\n",  "\n",  "from math import log2, floor\n",  "from random import randint\n",  "\n",  "def decompose_even_number(n):\n",  " '''Find the values r and d such that n == d * 2**r.'''\n",  " r = 0\n",  " d = n\n",  " for i in range(1, floor(log2(n)) + 1):\n",  " if (d/2).is_integer():\n",  " d = d/2\n",  " r += 1\n",  " else:\n",  " break\n",  " return (r, int(d))\n",  "\n",  "verbose = False # Print during execution\n",  "\n",  "def miller_rabin_prime(n):\n",  " ''':param n: should be odd, and n>3. No primes can be even, for obvious reasons.'''\n",  " accuracy = 3 # How many different a to test\n",  " r, d = decompose_even_number(n-1) # n is odd, thus n-1 is even\n",  " verbose and print(f\"Testing {n} a maximum of {accuracy} times...\")\n",  " \n",  " for _ in range(accuracy):\n",  " try:\n",  " a = randint(2, n-2) # The \"witness\". TODO: don't reuse a witness.\n",  " verbose and print(f\"\\tTesting {n} with a={a}\")\n",  "\n",  " x = (a**d) % n\n",  " if x == 1 or x == n-1:\n",  " verbose and print(f\"\\t({a}**{d})%{n} was 1 or -1 already!\")\n",  " continue # This witness was immediately 1 or -1 (aka n-1), so it can't help us any more.\n",  " for squaring in range(r-1):\n",  " x = (x**2) % n\n",  " verbose and print(f\"\\tx={x}\")\n",  " if x == n-1:\n",  " raise Exception(\"Try next witness\") # This witness can't help any more.\n",  " return \"composite\" # The witness didnt uphold the criteria for a prime, so n is composite!\n",  " except Exception as tryNext:\n",  " continue\n",  " return \"probably prime\"\n",  " \n",  "s, d = decompose_even_number(168)\n",  "print(f\"168 = {d}*2^{s} = {d*(2**s)}\")\n",  "s, d = decompose_even_number(65536)\n",  "print(f\"65536 = {d}*2^{s} = {d*(2**s)}\")\n",  "print()\n",  "\n",  "for value in [16127, 561, 2465, 65537]:\n",  " print(f\"{value} is {miller_rabin_prime(value)}.\")\n"  ]  },  {  "cell_type": "markdown",  "metadata": {},  "source": [  "# Discrete logarithm problem\n",  "Given a prime p and an integer y with 0 < y < p, find x such that \n",  "\n",  "> $y = g^{x}\\mod{p}$\n",  "\n",  "No polynomial algorithims exists, but \"Baby-step giant-step\" is a reasonable one. \n",  "\n",  "Sources:\n",  "- https://stackoverflow.com/a/1832648/5976426\n",  "- https://en.wikipedia.org/wiki/Baby-step_giant-step\n",  "- https://gist.github.com/0xTowel/b4e7233fc86d8bb49698e4f1318a5a73\n",  "\n",  "### Square and multiply\n",  "\n",  "Used to simplify doing large exponentiations. Instead of solving $3^5$ as $3*3*3*3*3$, \n",  "\n",  "1. Convert the exponent to Binary.\n",  "2. For the first $1$, simply list the number\n",  "3. For each ensuing $0$, do Square operation\n",  "4. For each ensuing $1$, do Square and Multiply operations\n",  "\n",  "Example:\n",  "\n",  "5 = 101 in Binary\n",  "1 First One lists Number 3\n",  "0 Zero calls for Square (3)²\n",  "1 One calls for Square + Multiply ((3)²)²*3\n",  "\n",  "\n",  "https://www.practicalnetworking.net/stand-alone/square-and-multiply/\n"  ]  },  {  "cell_type": "code",  "execution_count": 12,  "metadata": {},  "outputs": [  {  "name": "stdout",  "output_type": "stream",  "text": [  "7894352216^(1) * 7894352216^(2) * 7894352216^(16) * 7894352216^(64) * 7894352216^(1024) * 7894352216^(8192) * 7894352216^(131072) * 7894352216^(2097152) * 7894352216^(33554432) * 7894352216^(67108864)\n",  "squarings: 26\n",  "mults: 9\n",  "\n",  "z = 7894352216^102900819 mod(604604729)\n",  "z = 355407489\n",  "\n",  "7894352216^102900819 mod(604604729) = 355407489\n"  ]  }  ],  "source": [  "# Square and multiply\n",  "from math import log, floor\n",  "\n",  "def square_and_multiply(y,e,n, verbose=False):\n",  " # prep\n",  " e_bin = bin(e)[:1:-1] # e0, e1, e2, e3, ... , ek\n",  " k = len(e_bin)\n",  " z = 1\n",  " yi = y\n",  " indices = []\n",  "\n",  " for i in range(k):\n",  " ei = int(e_bin[i])\n",  " if ei == 1:\n",  " z = z*yi % n\n",  " if ei < k:\n",  " yi = yi*yi % n\n",  " if ei:\n",  " indices.append(i)\n",  "\n",  " if verbose:\n",  " mults = str(e_bin).count(\"1\") - 1\n",  " squarings = floor( log(e, 2) )\n",  "\n",  " print(f\"{' * '.join([f'{y}^({2**i})' for i in indices])}\")\n",  " print(f\"squarings: {squarings}\")\n",  " print(f\"mults: {mults}\")\n",  " print()\n",  " print(f\"z = {y}^{e} mod({n})\")\n",  " print(f\"z = {z}\\n\")\n",  " return z\n",  "\n",  "# y^e mod(n)\n",  "y = 7894352216\n",  "e = 102900819\n",  "n = 604604729\n",  "\n",  "z = square_and_multiply(y,e,n, verbose=True)\n",  "\n",  "print(f\"{y}^{e} mod({n}) = {z}\")"  ]  },  {  "cell_type": "code",  "execution_count": 13,  "metadata": {},  "outputs": [  {  "name": "stdout",  "output_type": "stream",  "text": [  "x = 102900819\n",  "y = g^x mod p ==> 355407489 = 7894352216^102900819 mod 604604729\n",  "Test: True\n"  ]  }  ],  "source": [  "from math import ceil, sqrt\n",  "\n",  "def bsgs(g, y, p):\n",  " '''\n",  " Solve for x in y = g^x mod p given a prime p.\n",  " If p is not prime, you shouldn't use BSGS anyway.\n",  " '''\n",  " N = ceil(sqrt(p - 1)) # phi(p) is p-1 if p is prime\n",  "\n",  " # Store hashmap of g^{1...m} (mod p). Baby step.\n",  " tbl = {pow(g, i, p): i for i in range(N)}\n",  "\n",  " # Precompute via Fermat's Little Theorem\n",  " c = pow(g, N * (p - 2), p)\n",  "\n",  " # Search for an equivalence in the table. Giant step.\n",  " for j in range(N):\n",  " h = (y * pow(c, j, p)) % p\n",  " if h in tbl:\n",  " return j * N + tbl[h]\n",  "\n",  " # Solution not found\n",  " return None\n",  "\n",  "def disc_log_test(y, g, x, p):\n",  " #return y == (g**x) % p\n",  " return y == square_and_multiply(g,x,p,False)\n",  "\n",  "\n",  "# y = g^x mod p\n",  "y = 355407489\n",  "g = 7894352216\n",  "p = 604604729 # Must be prime\n",  "\n",  "#y = 34\n",  "#g = 7324\n",  "#p = 4363 # Must be prime\n",  "\n",  "x = bsgs(g, y, p)\n",  "\n",  "print(f\"x = {x}\")\n",  "\n",  "print(f\"y = g^x mod p ==> {y} = {g}^{x} mod {p}\")\n",  "\n",  "print(f\"Test: {disc_log_test(y,g,x,p)}\")"  ]  },  {  "cell_type": "markdown",  "metadata": {},  "source": [  "# RSA\n",  "\n",  "### Key generation\n",  "\n",  "1. Let $p$ and $q$ be distinct prime numbers, randomly chosen from the set of all prime numbers of a certain size.\n",  "2. Compute $n = pq$.\n",  "3. Select $e$ randomly with $gcd(e, \\varphi(n)) = 1$.\n",  "4. Compute $d = e−1 \\mod \\varphi(n)$.\n",  "5. The public key is the pair $n$ and $e$.\n",  "6. The private key consists of the values $p$, $q$ and $d$.\n",  "\n",  "### Encryption\n",  "\n",  "The public key for encryption is $KE = (n, e)$\n",  "1. Input is any value $M$ where $0 < M < n$. \n",  "2. Compute $C = E(M,KE) = Me mod n$.\n",  "\n",  "### Decryption\n",  "\n",  "The private key for decryption is KD = d (values p and q are not used here).\n",  "1. Compute D(C,KD) = Cd mod n = M."  ]  },  {  "cell_type": "code",  "execution_count": 24,  "metadata": {},  "outputs": [  {  "name": "stdout",  "output_type": "stream",  "text": [  "p=43, q=59, n=2537, 𝜑(n)=2436, e=5, d=1949\n",  "\n",  "Public key: n=2537, e=5\n",  "Private key: p=43, q=59, d=1949\n",  "\n",  "{'n': 2537, 'e': 5, 'p': 43, 'q': 59, 'd': 1949}\n"  ]  }  ],  "source": [  "import random\n",  "\n",  "def rsa_keygen(p=None, q=None, e=None, prime_range=1000, verbose=False):\n",  " # Generate a set of primes. In practice, this should be\n",  " # a large number of large primes.\n",  " primes = [i for i in range(0,prime_range) if is_prime(i)]\n",  "\n",  " if not p: p = random.choice(primes)\n",  " if not q: q = random.choice(primes)\n",  " n = p*q\n",  "\n",  " # Euler function 𝜑(n) \n",  " phi_n = totient(n)\n",  "\n",  " # Select e randomly with gcd(e, φ(n)) = 1.\n",  " if not e: e = random.randint(1, phi_n)\n",  " while not gcd(e, phi_n) == 1:\n",  " e = random.randint(1,phi_n)\n",  "\n",  " # Compute d = e−1 mod φ(n).\n",  " d = inverse(e, phi_n)\n",  " \n",  " if verbose:\n",  " print(f\"p={p}, q={q}, n={n}, 𝜑(n)={phi_n}, e={e}, d={d}\\n\")\n",  " print(f\"Public key: n={n}, e={e}\")\n",  " print(f\"Private key: p={p}, q={q}, d={d}\\n\")\n",  "\n",  " return {\"n\":n, \"e\":e, \"p\":p, \"q\":q, \"d\":d}\n",  "\n",  "# Example from slides\n",  "rsa_keys = rsa_keygen(p=43, q=59, e=5, verbose=True)\n",  "\n",  "# Random keys\n",  "#rsa_keys = rsa_keygen(verbose=True)\n",  "print(rsa_keys)"  ]  },  {  "cell_type": "code",  "execution_count": 22,  "metadata": {},  "outputs": [  {  "name": "stdout",  "output_type": "stream",  "text": [  "C = M^e (mod n) = 50^102900819 (mod 604604729) = 2488\n",  "\n",  "Ciphertext: 2488\n"  ]  }  ],  "source": [  "def rsa_encrypt(M, rsa_keys):\n",  " # e and n are the public key\n",  " e = rsa_keys['e']\n",  " n = rsa_keys['n']\n",  "\n",  " # C = M^e (mod n)\n",  " return square_and_multiply(M,e,n)\n",  "\n",  "M = 50\n",  "C = rsa_encrypt(M, rsa_keys)\n",  "print(f\"C = M^e (mod n) = {M}^{e} (mod {n}) = {C}\\n\")\n",  "print(f\"Ciphertext: {C}\")"  ]  },  {  "cell_type": "code",  "execution_count": 23,  "metadata": {},  "outputs": [  {  "name": "stdout",  "output_type": "stream",  "text": [  "M = C^d (mod n) = 2488^1 (mod 604604729) = 50\n",  "\n",  "Message: 50\n"  ]  }  ],  "source": [  "def rsa_decrypt(C, rsa_keys):\n",  " # private key\n",  " p = rsa_keys['p']\n",  " q = rsa_keys['q']\n",  " d = rsa_keys['d']\n",  " n = p*q\n",  "\n",  " # M = C^d (mod n)\n",  " return square_and_multiply(C,d,n)\n",  "\n",  "M = rsa_decrypt(C, rsa_keys)\n",  "print(f\"M = C^d (mod n) = {C}^{d} (mod {n}) = {M}\\n\")\n",  "print(f\"Message: {M}\")"  ]  }  ],  "metadata": {  "kernelspec": {  "display_name": "Python 3",  "language": "python",  "name": "python3"  },  "language_info": {  "codemirror_mode": {  "name": "ipython",  "version": 3  },  "file_extension": ".py",  "mimetype": "text/x-python",  "name": "python",  "nbconvert_exporter": "python",  "pygments_lexer": "ipython3",  "version": "3.7.3"  }  },  "nbformat": 4,  "nbformat_minor": 4 }